jmc

Let $a_1=a$ and $a_2=b.$ Then $$\begin{array}{rl}{a}_3& =a+b,\\ a_4& =a+2b,\\ a_5& =2a+3b,\\ a_6& =3a+5b,\\ a_7& =5a+8b,\\ a_8& =8a+13b.\end{array}$$Hence, $5a+8b=120.$ Then $5a=120-8b=8(15-b).$ Since 5 is relatively prime to 8, $a$ is divisible by 8.

If $a=8,$ then $b=10.$ If $a=16,$ then $b=5,$ which does not work, because the sequence is increasing, so $b>a.$ Note that higher values of $b$ return lower values of $a,$ so the only possible value of $a$ is 8. Then $b=10,$ so $a_8=8a+13b=\boxed{194}.$